Divide the following complex numbers. $ \dfrac{-9+19i}{-1+5i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1-5i}$ $ \dfrac{-9+19i}{-1+5i} = \dfrac{-9+19i}{-1+5i} \cdot \dfrac{{-1-5i}}{{-1-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-9+19i) \cdot (-1-5i)} {(-1+5i) \cdot (-1-5i)} = \dfrac{(-9+19i) \cdot (-1-5i)} {(-1)^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-9+19i) \cdot (-1-5i)} {(-1)^2 - (5i)^2} = $ $ \dfrac{(-9+19i) \cdot (-1-5i)} {1 + 25} = $ $ \dfrac{(-9+19i) \cdot (-1-5i)} {26} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-9+19i}) \cdot ({-1-5i})} {26} = $ $ \dfrac{{-9} \cdot {(-1)} + {19} \cdot {(-1) i} + {-9} \cdot {-5 i} + {19} \cdot {-5 i^2}} {26} $ Evaluate each product of two numbers. $ \dfrac{9 - 19i + 45i - 95 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{9 - 19i + 45i + 95} {26} = \dfrac{104 + 26i} {26} = 4+i $